Outer measure explained

In the mathematical field of measure theory, an outer measure or exterior measure is a function defined on all subsets of a given set with values in the extended real numbers satisfying some additional technical conditions. The theory of outer measures was first introduced by Constantin Carathéodory to provide an abstract basis for the theory of measurable sets and countably additive measures. Carathéodory's work on outer measures found many applications in measure-theoretic set theory (outer measures are for example used in the proof of the fundamental Carathéodory's extension theorem), and was used in an essential way by Hausdorff to define a dimension-like metric invariant now called Hausdorff dimension. Outer measures are commonly used in the field of geometric measure theory.

Measures are generalizations of length, area and volume, but are useful for much more abstract and irregular sets than intervals in

R

or balls in

R3

. One might expect to define a generalized measuring function

\varphi

on

R

that fulfills the following requirements:
  1. Any interval of reals

[a,b]

has measure

b-a

  1. The measuring function

\varphi

is a non-negative extended real-valued function defined for all subsets of

R

.
  1. Translation invariance: For any set

A

and any real

x

, the sets

A

and

A+x=\{a+x:a\inA\}

have the same measure

(Aj)

of pairwise disjoint subsets of

R

infty
\varphi\left(cup
i=1

Ai\right)=

infty
\sum
i=1

\varphi(Ai).

It turns out that these requirements are incompatible conditions; see non-measurable set. The purpose of constructing an outer measure on all subsets of

X

is to pick out a class of subsets (to be called measurable) in such a way as to satisfy the countable additivity property.

Outer measures

Given a set

X,

let

2X

denote the collection of all subsets of

X,

including the empty set

\varnothing.

An outer measure on

X

is a set function\mu: 2^X \to [0, \infty]such that

\mu(\varnothing)=0

A,B1,B2,\ldots

of

X,

\text A \subseteq \bigcup_^\infty B_j \text \mu(A) \leq \sum_^\infty \mu(B_j).Note that there is no subtlety about infinite summation in this definition. Since the summands are all assumed to be nonnegative, the sequence of partial sums could only diverge by increasing without bound. So the infinite sum appearing in the definition will always be a well-defined element of

[0,infty].

If, instead, an outer measure were allowed to take negative values, its definition would have to be modified to take into account the possibility of non-convergent infinite sums.

An alternative and equivalent definition.[1] Some textbooks, such as Halmos (1950), instead define an outer measure on

X

to be a function

\mu:2X\to[0,infty]

such that

\mu(\varnothing)=0

A

and

B

are subsets of

X

with

A\subseteqB,

then

\mu(A)\leq\mu(B)

B1,B2,\ldots

of

X,

\mu\left(\bigcup_^\infty B_j\right) \leq \sum_^\infty \mu(B_j).
Proof of equivalence.
Suppose that

\mu

is an outer measure in sense originally given above. If

A

and

B

are subsets of

X

with

A\subseteqB,

then by appealing to the definition with

B1=B

and

Bj=\varnothing

for all

j\geq2,

one finds that

\mu(A)\leq\mu(B).

The third condition in the alternative definition is immediate from the trivial observation that

\cupjBj\subseteq\cupjBj.

Suppose instead that

\mu

is an outer measure in the alternative definition. Let

A,B1,B2,\ldots

be arbitrary subsets of

X,

and suppose that A \subseteq \bigcup_^\infty B_j.One then has\mu(A) \leq \mu\left(\bigcup_^\infty B_j\right) \leq \sum_^\infty\mu(B_j),with the first inequality following from the second condition in the alternative definition, and the second inequality following from the third condition in the alternative definition. So

\mu

is an outer measure in the sense of the original definition.

Measurability of sets relative to an outer measure

Let

X

be a set with an outer measure

\mu.

One says that a subset

E

of

X

is

\mu

-measurable
(sometimes called Carathéodory-measurable relative to

\mu

, after the mathematician Carathéodory) if and only if\mu(A) = \mu(A \cap E) + \mu(A \setminus E)for every subset

A

of

X.

Informally, this says that a

\mu

-measurable subset is one which may be used as a building block, breaking any other subset apart into pieces (namely, the piece which is inside of the measurable set together with the piece which is outside of the measurable set). In terms of the motivation for measure theory, one would expect that area, for example, should be an outer measure on the plane. One might then expect that every subset of the plane would be deemed "measurable," following the expected principle that\operatorname(A \cup B) = \operatorname(A) + \operatorname(B)whenever

A

and

B

are disjoint subsets of the plane. However, the formal logical development of the theory shows that the situation is more complicated. A formal implication of the axiom of choice is that for any definition of area as an outer measure which includes as a special case the standard formula for the area of a rectangle, there must be subsets of the plane which fail to be measurable. In particular, the above "expected principle" is false, provided that one accepts the axiom of choice.

The measure space associated to an outer measure

It is straightforward to use the above definition of

\mu

-measurability to see that

A\subseteqX

is

\mu

-measurable then its complement

X\setminusA\subseteqX

is also

\mu

-measurable.The following condition is known as the "countable additivity of

\mu

on measurable subsets."

A1,A2,\ldots

are

\mu

-measurable pairwise-disjoint (

Ai\capAj=\emptyset

for

ij

) subsets of

X

, then one has \mu\Big(\bigcup_^\infty A_j\Big) = \sum_^\infty\mu(A_j).
Proof of countable additivity.
One automatically has the conclusion in the form "

\leq

" from the definition of outer measure. So it is only necessary to prove the "

\geq

" inequality. One has \mu\Big(\bigcup_^\infty A_j\Big)\geq\mu\Big(\bigcup_^N A_j\Big) for any positive number

N,

due to the second condition in the "alternative definition" of outer measure given above. Suppose (inductively) that \mu\Big(\bigcup_^ A_j\Big)=\sum_^\mu(A_j)Applying the above definition of

\mu

-measurability with

A=A1\cup\cupAN

and with

E=AN,

one has \begin\mu\Big(\bigcup_^N A_j\Big) &= \mu\left(\Big(\bigcup_^N A_j\Big)\cap A_N\right) + \mu\left(\Big(\bigcup_^N A_j\Big)\smallsetminus A_N\right) \\&= \mu(A_N) + \mu\Big(\bigcup_^A_j\Big)\end which closes the induction. Going back to the first line of the proof, one then has \mu\Big(\bigcup_^\infty A_j\Big)\geq\sum_^N \mu(A_j) for any positive integer

N.

One can then send

N

to infinity to get the required "

\geq

" inequality.
A similar proof shows that:

A1,A2,\ldots

are

\mu

-measurable subsets of

X,

then the union
infty
cup
i=1

Ai

and intersection
infty
cap
i=1

Ai

are also

\mu

-measurable.

The properties given here can be summarized by the following terminology:One thus has a measure space structure on

X,

arising naturally from the specification of an outer measure on

X.

This measure space has the additional property of completeness, which is contained in the following statement:

A\subseteqX

such that

\mu(A)=0

is

\mu

-measurable.This is easy to prove by using the second property in the "alternative definition" of outer measure.

Restriction and pushforward of an outer measure

Let

\mu

be an outer measure on the set

X

.

Pushforward

Given another set

Y

and a map

f:X\toY

define

f\sharp\mu:2Y\to[0,infty]

by
-1
(f
\sharp\mu)(A)=\mu(f

(A)).

One can verify directly from the definitions that

f\sharp\mu

is an outer measure on

Y

.

Restriction

Let be a subset of . Define by

\muB(A)=\mu(A\capB).

One can check directly from the definitions that is another outer measure on .

Measurability of sets relative to a pushforward or restriction

If a subset of is -measurable, then it is also -measurable for any subset of .

Given a map and a subset of, if is -measurable then is -measurable. More generally, is -measurable if and only if is -measurable for every subset of .

Regular outer measures

Definition of a regular outer measure

Given a set, an outer measure on is said to be regular if any subset

A\subseteqX

can be approximated 'from the outside' by -measurable sets. Formally, this is requiring either of the following equivalent conditions:

\mu(A)=inf\{\mu(B)\midA\subseteqB,Bisμ-measurable\}

\mu(B)=\mu(A)

.It is automatic that the second condition implies the first; the first implies the second by taking the countable intersection of

Bi

with

\mu(Bi)\to\mu(A)

The regular outer measure associated to an outer measure

Given an outer measure on a set, define by

\nu(A)=inf\{\mu(B):\mu-measurablesubsetsB\subsetXwithB\supsetA\}.

Then is a regular outer measure on which assigns the same measure as to all -measurable subsets of . Every -measurable subset is also -measurable, and every -measurable subset of finite -measure is also -measurable.

So the measure space associated to may have a larger σ-algebra than the measure space associated to . The restrictions of and to the smaller σ-algebra are identical. The elements of the larger σ-algebra which are not contained in the smaller σ-algebra have infinite -measure and finite -measure.

From this perspective, may be regarded as an extension of .

Outer measure and topology

Suppose is a metric space and an outer measure on . If has the property that

\varphi(E\cupF)=\varphi(E)+\varphi(F)

whenever

d(E,F)=inf\{d(x,y):x\inE,y\inF\}>0,

then is called a metric outer measure.

Theorem. If is a metric outer measure on, then every Borel subset of is -measurable. (The Borel sets of are the elements of the smallest -algebra generated by the open sets.)

Construction of outer measures

See also: Valuation (measure theory).

There are several procedures for constructing outer measures on a set. The classic Munroe reference below describes two particularly useful ones which are referred to as Method I and Method II.

Method I

Let be a set, a family of subsets of which contains the empty set and a non-negative extended real valued function on which vanishes on the empty set.

Theorem. Suppose the family and the function are as above and define

\varphi(E)=infl\{

infty
\sum
i=0

p(Ai)|E\subseteqcup

infty
i=0

Ai,\foralli\inN,Ai\inCr\}.

That is, the infimum extends over all sequences of elements of which cover, with the convention that the infimum is infinite if no such sequence exists. Then is an outer measure on .

Method II

The second technique is more suitable for constructing outer measures on metric spaces, since it yields metric outer measures. Suppose is a metric space. As above is a family of subsets of which contains the empty set and a non-negative extended real valued function on which vanishes on the empty set. For each, let

C\delta=\{A\inC:\operatorname{diam}(A)\leq\delta\}

and

\varphi\delta(E)=infl\{

infty
\sum
i=0

p(Ai)|E\subseteqcup

infty
i=0

Ai,\foralli\inN,Ai\inC\deltar\}.

Obviously, when since the infimum is taken over a smaller class as decreases. Thus

\lim\delta\varphi\delta(E)=\varphi0(E)\in[0,infty]

exists (possibly infinite).

Theorem. is a metric outer measure on .

This is the construction used in the definition of Hausdorff measures for a metric space.

See also

References

. Constantin Carathéodory. Vorlesungen über reelle Funktionen. 3rd. 1918. 1968. German. Chelsea Publishing. 978-0828400381.

. Geometric Measure Theory. Classics in Mathematics. 1st ed reprint. Berlin, Heidelberg, New York. 1969. 1996. 978-3540606567. Springer Verlag. Herbert Federer.

. Paul Halmos. Measure theory. Springer Verlag. Berlin, Heidelberg, New York. Graduate Texts in Mathematics. 1978. 978-0387900889. 2nd. 1950.

External links

Notes and References

  1. The original definition given above follows the widely cited texts of Federer and of Evans and Gariepy. Note that both of these books use non-standard terminology in defining a "measure" to be what is here called an "outer measure."